Example Consider three point charges located at the corners of a right triangle as shown in the figure, where q1 = q3 = 5.0 µC, q2 = 2.0 µC, and a = 0.10 m. Find the electric field at q3 and the resultant force exerted on this charge. At the location of q3, the total electric field E is equal to the vector sum of the two individual fields, E, and E2. First, calculate the magnitude of E, and E2. 92 The distance between q, and q3 can be calculated using the Pythagorean Theorem: a r13 = Va? + a² = VZa 91 E = k r132 m2 5.0 x 10-6 C C2) 2(0.10 m)² 8.988 x 10° N - = 2.2 x 106 N/C 92 2.0 x 10-6 C E2 = k- r232 8.988 x 10° N C2 (0.10 m)? = 1.8 x 106 N/c Find the components of the resultant electric field at the location of q3. E, = E1x + E2x = 1.6 x 10° N/C – 1.8 × 10° N/C = -0.2 × 10° N/C %3D Ey = E1y + E2y = 1.6 × 106 N/C

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Consider three point charges located at the corners of a right triangle as shown in the
figure, where q1 = q3 = 5.0 µC, q2 = 2.0 µC, and a = 0.10 m. Find the electric field at q3
and the resultant force exerted on this charge.
At the location of q3, the total electric field E is equal to
the vector sum of the two individual fields, E, and E2. First,
calculate the magnitude of E, and E2.
92
The distance between q, and q3 can be calculated using
the Pythagorean Theorem:
a
r13 = Va? + a² =
VZa
91
E = k
r132
m2 5.0 x 10-6 C
C2) 2(0.10 m)²
8.988 x 10° N -
= 2.2 x 106 N/C
92
2.0 x 10-6 C
E2 = k-
r232
8.988 x 10° N
C2
(0.10 m)?
= 1.8 x 106 N/c
Transcribed Image Text:Example Consider three point charges located at the corners of a right triangle as shown in the figure, where q1 = q3 = 5.0 µC, q2 = 2.0 µC, and a = 0.10 m. Find the electric field at q3 and the resultant force exerted on this charge. At the location of q3, the total electric field E is equal to the vector sum of the two individual fields, E, and E2. First, calculate the magnitude of E, and E2. 92 The distance between q, and q3 can be calculated using the Pythagorean Theorem: a r13 = Va? + a² = VZa 91 E = k r132 m2 5.0 x 10-6 C C2) 2(0.10 m)² 8.988 x 10° N - = 2.2 x 106 N/C 92 2.0 x 10-6 C E2 = k- r232 8.988 x 10° N C2 (0.10 m)? = 1.8 x 106 N/c
Find the components of the resultant electric field at the location of q3.
E, = E1x + E2x = 1.6 x 10° N/C – 1.8 × 10° N/C = -0.2 × 10° N/C
%3D
Ey = E1y + E2y = 1.6 × 106 N/C
Transcribed Image Text:Find the components of the resultant electric field at the location of q3. E, = E1x + E2x = 1.6 x 10° N/C – 1.8 × 10° N/C = -0.2 × 10° N/C %3D Ey = E1y + E2y = 1.6 × 106 N/C
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